If a rectangular field of an area $100$ square units is divided into two parts of area $36$ square units and $64$ square units along the length such that the length of the original field and both the newly created fields are coprime to each other, then the sum of the length of original field and both the newly created fields is

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Solution

$100 = 2^{2} \times 5^{2}$
$64 = 2^{6}$
$36 = 2^{2} \times 3^{2}$
Since, the common factor among all is $2^{2}$
So, the breadth is $2^{2} = 4$ and the lengths are
$100 = 4 \times 25$
$64 = 4 \times 16$
$36 = 4 \times 9$
So, sum is $25 + 16 + 9 = 50$ units

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100=22×52 64=26 36=22×32 Since, the common factor among all is 22 So, the breadth is 22=4 and the lengths are 100=4×25 64=4×16 36=4×9 So, sum is 25+16+9=50 units

$100 = 2^{2} \times 5^{2}$
$64 = 2^{6}$
$36 = 2^{2} \times 3^{2}$
Since, the common factor among all is $2^{2}$
So, the breadth is $2^{2} = 4$ and the lengths are
$100 = 4 \times 25$
$64 = 4 \times 16$
$36 = 4 \times 9$
So, sum is $25 + 16 + 9 = 50$ units